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Example: The half-cell reaction: 2Fe2+ + 3H2O = Fe2O3(s) + 6H+ + 2e

This half-cell reaction can be decomposed into three simpler reactions, only two of which are found in Table 10.1:

A^ o(Fe/Fe2+) = -2x96.5x0.44 = -85 kJ/mole A^(Fe/Fe2O3) = -120x4.184 = 502 kJ/mole A^o(H2O/O2) = -4x96.5 x(-1.23) = 475 kJ/mole

The first and last of these half-cell reactions are numbers 7 and 1, respectively, in Table 10.1. The middle reaction does not involve either aqueous ions or electrons; its free energy change (at 25oC) is obtained from the Ellingham diagram of Fig. 10.3. The free energy change of the desired half-cell reaction is obtained by algebraically combining the above three component reactions:

A^°(Fe2+/Fe2O3) = 3/2 A^°(H2O/O2) + 3/2 A^°(Fe/Fe2O3) - 2 A^°(Fe/Fe2+) = 129 kJ/mole Converting this to the half-cell EMF by use of Eq (10.30) yields:

Problem 10.2 is another example of combining entries in Table 10.1 to obtain the standard electrode potential of a reaction not listed in the table.

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